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3 years ago

Compare the purple and orange arrows. Does more energy go into breaking bonds, or is more energy released when new bonds form?

Chemistry

1 answer:

Nastasia [14]3 years ago

3 0

Answer:

The breaking of chemical bonds never releases energy to the external environment. Energy is only released when chemical bonds are formed. In general, a chemical reaction involves two steps: 1) the original chemical bonds between the atoms are broken, and 2) new bonds are formed. These two steps are sometimes lumped into one event for simplicity, but they are really two separate events. For instance, when you burn methane (natural gas) in your stove, the methane is reacting with oxygen to form carbon dioxide and water. Chemists often write this as:

CH4 + 2 O2 → CO2 + 2 H2O + energy

Explanation:

here is the answer

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Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0

grandymaker [24]

The given question is incomplete. The complete question is:

Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. Determine the mass of glucose (C6H1206) produced

Answer: 60.0 g of glucose

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $CO_2$

$\text{Number of moles}=\frac{88.0g}{44g/mol}=2.0moles$

b) moles of $H_2O$

$\text{Number of moles}=\frac{64.0g}{18g/mol}=3.5moles$

$6CO_2+6H_2O\rightarrow C_{6}H_{12}O_6+6O_2$

According to stoichiometry :

6 moles of $CO_2$ require = 6 moles of $H_2O$

Thus 2.0 moles of $CO_2$ require= $\frac{6}{6}\times 2.0=2.0moles$ of $H_2O$

Thus $CO_2$ is the limiting reagent as it limits the formation of product.

As 6 moles of $CO_2$ give = 1 moles of glucose

Thus 2.0 moles of $CO_2$ give = $\frac{1}{6}\times 2.0=0.33moles$ of glucose

Mass of glucose = $moles\times {\text {Molar mass}}=0.33moles\times 180g/mol=60g$

Thus 60.0 g of glucose will be produced from 88.0 g of carbon dioxide and 64.0 g of water

8 0

3 years ago

Is milk a mixture, solution, or both?

lara [203]

It is both. It is a mixture because it has two or more things mixed in it. It is a solution, because all of the components and things mixed in the milk are uniform and milk isn't chunky or discolored because of that.
Hope this helps.

5 0

3 years ago

Read 2 more answers

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 × 10 − 5), with 0.1000 M NaOH

Zina [86]

Here is the full question

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 × 10 − 5), with 0.1000 M NaOH solution after the following additions of titrant (total volume of added base given):

a) 10.00 mL

pH =

b) 20.10 mL

pH =

c) 25.00 mL

pH =

Answer:

pH = 4.81

pH = 10.40

pH = 12.04

Explanation:

Number of moles of butanoic acid

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $10.00 \ mL * \frac{L}{1000 \ mL }* \frac{0.1000 \ mol }{L}$

= 0.001000 mol

pKa of butanoic acid = - log Ka

= - log ( 1.54 × 10⁻⁵)

= 4.81

Equation for the reaction is expressed as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

The ICE Table is expressed as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

Initial 0.002000 0.001000 0

Change - 0.001000 - 0.001000 + 0.001000

Equilibrium 0.001000 0 0.001000

Total Volume = (20.00 + 10.00 ) mL

= 30.00 mL = 0.03000 L

Concentration of [CH₃CH₂CH₂COOH] = $\frac{0.001000 \ mol}{ 0.03000 \ L }$

= 0.03333 M

Concentration of [CH₃CH₂COO⁻] = $\frac{0.001000 \ mol}{ 0.03000 \ L}$

= 0.03333 M

By Henderson- Hasselbalch equation

pH = pKa + log $\frac{conjugate \ base}{acid }$

pH = pKa + log $\frac{CH_3CH_2CH_2COO^-}{CH_3CH_2CH_2COOH}$

PH = 4.81 + log $\frac{0.03333}{0.03333}$

pH = 4.81

Thus; the pH of the resulted buffer solution after 10.00 mL of NaOH was added = 4.81

b )

After the equivalence point, we all know that the pH of the solution will now definitely be determined by the excess H⁺

Number of moles of butanoic acid

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $20.10 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002010 mol

Following the previous equation of reaction , The ICE Table for this process is as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

Initial 0.002000 0.002010 0

Change - 0.002000 -0.002000 + 0.002000

Equilibrium 0 0.000010 0.002000

We can see here that the base is present in excess;

NOW, number of moles of base present in excess

= ( 0.002010 - 0.002000) mol

= 0.000010 mol

Total Volume = (20.00 + 20.10 ) mL

= 40.10 mL × $\frac{1 \ L}{1000 \ mL }$

= 0.04010 L

Concentration of acid [OH⁻] = $\frac{0.000010 \ mol}{0.04010 \ L }$

= $2.494*10^{-4} M$

Using the ionic product of water:

$[H_3O^+] = \frac{K \omega }{[OH^-]}$

where

$K \omega = 10^{-14}$

$[H_3O^+] = \frac{1.0*10^{-14}}{2.494*10^{-14}}$

= $4.0*10^{-11}M$

pH = - log $[H_3O^+}]$

pH = - log [ $4.0*10^{-11}M$ ]

pH = 10.40

Thus, the pH of the solution after the equivalence point = 10.40

After the equivalence point, pH of the solution is determined by the excess H⁺.

Number of moles of butanoic acid

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $25.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002500 mol

From our chemical equation; The ICE Table can be illustrated as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

Initial 0.002000 0.002500 0

Change - 0.002000 -0.002000 +0.002000

Equilibrium 0 0.000500 0.002000

Base is present in excess

Number of moles of base present in excess = [ 0.002500 - 0.002000] mol

= 0.000500 mol

Total Volume = ( 20.00 + 25.00 ) mL

= 45.00 mL

= 45.00 × $\frac{1 \ L}{1000 \ mL }$

= 0.04500 L

Concentration of acid [OH⁻] = $\frac{0.0005000 \ mol}{ 0.04500 \ L }$

= 0.01111 M

Using the ionic product of water $[H_3O^+] = \frac{K \omega }{[OH^+]}$

= $\frac{1.0*10^{-14}}{0.01111}$

= $9.0*10^{-13}$ M

pH = - log $[H_3O^+}]$

pH = - log [ $9.0*10^{-13}M$ ]

pH = 12.04

Thus, the pH of the solution after the equivalence point = 12.04

4 0

3 years ago

Science!

tresset_1 [31]

Box will accelerate to the left.

7 0

3 years ago

What is the density of a rectangular block that has a mass of 500 grams and a volume of 50cm^3

oee [108]

Density = mass / volume

D = 500 g / 50 cm³

D = 10 g/cm³

hope this helps!

6 0

4 years ago